Solving non linear 2nd order differential equation

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Reuben Salisbury
Reuben Salisbury 2020 年 4 月 7 日
コメント済み: Reuben Salisbury 2020 年 4 月 7 日
I am trying to solve the differential equation:
mx''+cx'+kx=ky+cy''
y is the input and x is the output.
y=Y0sin(wt)
where m=100, c=1300, k= 17000, Y0=input value, w=input value
initially I need to solve this, and then i need to plot displacement and velocity profiles.
clear
t=0:0.1:15; %time peroid
Y0= input('wave amplitude ') ; %Wave amplitude
l= input('length of wave ') ; %length of the wave
u=input('speed of boat'); %Boat Velocity
w=u/l; %frequency of wave
y=Y0*sin(w*t); %wave height model
Dr=0.5; %Required Damping Ratio
k=17000; %Spring Constant of suspension
m=100;
wn=(k/m)^0.5; %Natural Frequency
wd=wn*(1-Dr^2)^0.5; %Damped Natural Frequency
c=2*Dr*(k*m)^(0.5); %Damping coefficient
syms x(t)
ode=m*diff(x,t,2)+c*diff(x,t)+k*x==y+(2*Dr/wn)*diff(y,t);
xSol(t)=dsolve(ode);

採用された回答

Birdman
Birdman 2020 年 4 月 7 日
Try this:
clear
syms t
Y0= input('wave amplitude ') ; %Wave amplitude
l= input('length of wave ') ; %length of the wave
u=input('speed of boat'); %Boat Velocity
w=u/l; %frequency of wave
y=Y0*sin(w*t); %wave height model
Dr=0.5; %Required Damping Ratio
k=17000; %Spring Constant of suspension
m=100;
wn=(k/m)^0.5; %Natural Frequency
wd=wn*(1-Dr^2)^0.5; %Damped Natural Frequency
c=2*Dr*(k*m)^(0.5); %Damping coefficient
syms x(t)
Dx(t)=diff(x,t);
ode=m*diff(x,t,2)+c*diff(x,t)+k*x==y+(2*Dr/wn)*diff(y,t);
xSol(t)=dsolve(ode,[x(0)==0 Dx(0)==0]);%displacement
DxSol(t)=diff(xSol,t);%velocity
t=0:0.05:15;
plot(t,xSol(t),t,DxSol(t))
  1 件のコメント
Reuben Salisbury
Reuben Salisbury 2020 年 4 月 7 日
That's great, thank you so much!

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